2009-07-28 110 views
1

給定jQuery replaceWith Flash對象參數?

<div id="dialog" style="display:none; width:100%; vertical-align:middle;"> 
    <object id="player" classid="clsid:D27CDB6E-AE6D-11cf-96B8-444553540000" name="player" width="628" height="400"> 
     <param name="movie" value="../../help/videos/videoplayer.swf" /> 
     <param name="allowfullscreen" value="true" /> 
     <param name="allowscriptaccess" value="always" /> 
     <param name="flashvars" value="file=../../help/videos/video.flv" /> 
     <object type="application/x-shockwave-flash" data="../../help/videos/videoplayer.swf" width="628" height="400"> 
      <param name="movie" value="../../help/videos/videoplayer.swf" /> 
      <param name="allowfullscreen" value="true" /> 
      <param name="allowscriptaccess" value="always" /> 
      <param name="flashvars" value="file=../../help/videos/video.flv" /> 
     </object> 
    </object> 

爲什麼不會在

$("#dialog").find('<param name="flashvars" value="file=../../help/videos/video.flv" />').replaceWith('<param name="flashvars" value="file=../../help/videos/myfile.flv" />'); 

工作?

回答

3

你是不是傳遞一個選擇找到,你可以做這樣的事情:

$('#player param[name=flashvars]').attr('value','new value'); 
+1

不幸的是,這並不在IE瀏覽器。 – Rio 2009-07-28 19:25:46

0
var flashhtml = $("object").html(); 
    $('object').before("<div id='mydiv'>내 div</div>"); 
    $('object').remove(); 
    var flashattr = 'classid="clsid:D27CDB6E-AE6D-11cf-96B8-444553540000" codebase="http://download.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=6,0,29,0" width="583" height="297">'; 
    $('#mydiv').html('<object ' + flashattr + '<param name="wmode" value="transparent">' + flashhtml + '</object>');